# What does $\int_C V \,d\mathbf{l}$ mean?

What does $$\int_C V \,d\mathbf{l}$$ mean? I initially thought it was simply a line integral around $C$, that is, if $\mathbf{r}: [0,1] \longrightarrow \mathbb{R}^3$ is a paremetrization of $C$, then $$\int_C V d\mathbf{l} = \int_0^1 V(\mathbf{r}(t))\, \left|\frac{d}{dt}\mathbf{r}(t)\right| \, dt$$

But quoting my textbook (Field and Wave Electromagnetics, 2nd edition by David K. Cheng):

(...) $V$ is a scalar function of space, $d\mathbf{l}$ represents a differential increment of length, and $C$ is the path of integration. If the integration is to be carried out from a point $P_1$ to another point $P_2$, we write $\int_{P_1}^{P_2} V d\mathbf{l}$. If the integration is to be evalueated around a closed path $C$, we denote it by $\oint_C V d\mathbf{l}$. In Cartesian coordinates, it can be written as $$\int_C V d\mathbf{l} = \int_C V(x,y,z)[\mathbf{a}_x \,dx + \mathbf{a}_y \,dy + \mathbf{a}_z \,dz],$$ (...) which becomes $$\int_C V d\mathbf{l} = \mathbf{a}_x\int_C V(x,y,z)\,dx + \mathbf{a}_y\int_C V(x,y,z)\,dy + \mathbf{a}_z\int_C V(x,y,z)\,dz.$$ The three integrals on the right hand side are ordinary scalar integrals; they can be evaluated for a given $V(x,y,z)$ around a path $C$.

I don't understand what the integrals on the last expression mean. For example, in $$\int_C V(x,y,z) \, dx$$ how is this a an "ordinary scalar integral" if $C$ is a line that can possibly be outside of the $xx$ axis.

• Which textbook? Commented Apr 28, 2014 at 22:11
• What is the context? It looks like the result of such an integral will still be a vector, with both a magnitude and a direction.
– rob
Commented Apr 28, 2014 at 22:12
• The textbook is "Field and Wave Electromagnetics", 2nd edition by David K. Cheng (I changed the quote slightly...). Yes, it is supposeed to result in a vector, which is what I find strange. $d \mathbf{l}$ is a vector, the l is bold fonted. The author provides one example, I can copy it if it helps. Commented Apr 28, 2014 at 22:14
• And the context is a summary of vector analysis that will be used later on the book. Commented Apr 28, 2014 at 22:16

The differential element $dl$ is a projector over the path, and $a_i$ are the unitary vectors along the axis. In your formulation, the only mistake is the absolute value, as you begin with a vector, and ergo should get a vector.
Note that sometimes we want the area under a curve defined on a scalar field, $\int_C f(x,y,z) ds$. Note $ds$ is not a vector.
• Thank you. Okay, so if $\int_C f(x,y,z) \,ds$ is the area under the curve on a scalar field, what exactly is the difference between $\int_C f(x,y,z) \, dx$, $\int_C f(x,y,z) \, dy$, and $\int_C f(x,y,z) \, dz$? Commented Apr 28, 2014 at 23:15
• There, $dx$ is the $x$ component of $\dfrac{d\mathbf{r}}{dt} dt$, $dy$ is the $y$ component, and $dz$ is the $z$ component. Commented Apr 29, 2014 at 2:22